Respuesta :
Given:
• Mass, m = 0.166 kg
,• Initial velocity, u = 0.700 m/s
,• Time, t = 3.50 seconds
Let's solve for the following:
• (A). What is the average acceleration of the puck during this time?
To find the average acceleration, apply the motion formula:
[tex]v=u+at[/tex]Where:
• v is the final velocity = 0 m/s
,• u is the initial velocity = 0.700 m/s
,• a is the acceleration
,• t is the time = 3.50 seconds.
Plug in the values and solve for the acceleration, a.
[tex]\begin{gathered} 0=0.700+a(3.50) \\ \\ 0=0.700+3.50a \\ \\ 3.50a=-0.700 \end{gathered}[/tex]Divide both sides by 3.50:
[tex]\begin{gathered} \frac{3.50a}{3.50}=\frac{-0.700}{3.50} \\ \\ a=-0.2\text{ m/s}^2 \end{gathered}[/tex]Therefore, the acceleration of the puck during this time is -0.2 m/s.
The negative sign means the player was slowing down.
• (B). How far does it travel during this time.
To find the distance, apply the motion formula:
[tex]s=ut+\frac{1}{2}at^2[/tex]Where:
s is the distance travelled.
u is the initial velocity = 0.700 m/s
a is the acceleration = -0.2 m/s²
t is the time = 3.50 s
Thus, we have:
[tex]\begin{gathered} s=0.700(3.50)+\frac{1}{2}(-0.2)(3.50)^2 \\ \\ s=2.45-1.225 \\ \\ s=1.225\text{ m} \end{gathered}[/tex]Therefore, the distance traveled during this time is 1.225 meters.
• (C). Assume the puck stopped due to friction, calculate the coefficient of kinetic friction.
To calculate the coefficient of kinetic friction, we have:
[tex]ma=\mu mg[/tex]Where:
m is the mass = 0.166 kg
a is the acceleration = -0.2 m/s²
g is the acceleration due to gravity = 9.8 m/s²
μ is the coefficient of kinetic friction.
Thus, we have:
[tex]\begin{gathered} \mu=\frac{ma}{mg} \\ \\ \mu=\frac{a}{g} \\ \\ \mu=\frac{0.2}{9.8} \\ \\ u=0.0204 \end{gathered}[/tex]Therefore, the coefficient of kinetic friction 0.0204.
ANSWER:
• (A). -0.2 m/s²
• (B). 1.225 m
• (C). 0.0204
